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Three vertices of a Parallelogram ABCD

Question:

Three vertices of a Parallelogram ABCD are A (1, 2, 3), B (– 1, – 2, – 1) and C (2, 3, 2). Find the fourth vertex D.

Solution:

Given three consecutive vertices of a parallelogram $A B C D$ are $A(1,2,3), B(-1$, $-2,-1)$ and $C(2,3,2)$

Let the fourth vertex be $D(x, y, z)$.

Now by using midpoint formula,

Midpoint of diagonal $A C=P\left(\frac{1+2}{2}, \frac{2+3}{2}, \frac{3+2}{2}\right)=P\left(\frac{3}{2}, \frac{5}{2}, \frac{5}{2}\right)$

Midpoint of diagonal BD $=\mathrm{P}\left(\frac{\mathrm{x}+(-1)}{2}, \frac{\mathrm{y}+(-2)}{2}, \frac{\mathrm{z}+(-1)}{2}\right)=\mathrm{P}\left(\frac{3}{2}, \frac{5}{2}, \frac{5}{2}\right)$

On comparing we get

$\Rightarrow \frac{x-1}{2}=\frac{3}{2}$

$\Rightarrow x=4$

$\Rightarrow \frac{y-2}{2}=\frac{5}{2}$

$\Rightarrow y=7$

$\Rightarrow \frac{z-1}{2}=\frac{5}{2}$

$\Rightarrow z=6$

$\therefore \mathrm{D}(4,7,6)$ is the fourth vertex.

 

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